Problem: How many positive three-digit integers less than 500 have at least two digits that are the same?
Answer: Case 1: The last two digits of our integer are equal.  There are 10 possibilities for these last two digits and 4 choices for the hundred's digit, of a total of 40 possibilities.  (Note that this case includes 111, 222, 333, and 444.)

Case 2: The first two digits are equal and the third is different.  This occurs in $4\cdot 9 = 36$ ways, since we can choose the repeated digit in 4 ways and the remaining digit in 9.

Case 3:  The first and third digits are equal while the second is different.  This also occurs in 36 ways.

Thus we have a total of $40 + 36 + 36 = \boxed{112}$ integers.

OR

Another way to solve this problem is to find how many three-digit integers less than 500 have no digits that are the same. The first digit must be 1, 2, 3, or 4. The second digit can be any of the 9 digits not yet chosen, and the third digit can be any of the 8 digits not yet chosen, so there are a total of $4 \cdot 9 \cdot 8 = 288$ three-digit integers that have no digits that are the same and are less than 500. There are a total of $500 - 100 = 400$ three-digit integers that are less than 500, so we have a total of $400 - 288 = \boxed{112}$ integers that fit the problem.  (Solution by Alcumus user chenhsi.)